Determine the density of cesium chloride which crystallizes in bcc type structure with the edge length 412.1 pm. The atomic masses of Cs and Cl are 133 and 35.5 respectively.
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strcuture of Cesium chloride is body centre cubic type in which Cs+ placed within centre and Cl- placed at corner of lattice.
number of Cl- ions per unit cell = 8 × 1/8 = 1
number of Cs+ ions per unit cell = 1
so, there is only one molecule of CsCl in a lattice .
so, mass of one molecule of CsCl = molar mass of CsCl/Avogadro's number
= (133 + 35.5)/(6.023 × 10²³)
= 168.5/6.023 × 10²³
= 2.8 × 10^-22 g
volume of lattice = a³ , where a is edge length
= (412.1 × 10^-12)³
= 7 × 10^-23 cm³
so, density of CsCl crystal = 2.8 × 10^-22/7 × 10^-23
= 4 g/cm³
number of Cl- ions per unit cell = 8 × 1/8 = 1
number of Cs+ ions per unit cell = 1
so, there is only one molecule of CsCl in a lattice .
so, mass of one molecule of CsCl = molar mass of CsCl/Avogadro's number
= (133 + 35.5)/(6.023 × 10²³)
= 168.5/6.023 × 10²³
= 2.8 × 10^-22 g
volume of lattice = a³ , where a is edge length
= (412.1 × 10^-12)³
= 7 × 10^-23 cm³
so, density of CsCl crystal = 2.8 × 10^-22/7 × 10^-23
= 4 g/cm³
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